Subject: Chemistry — Electrochemistry
Question: Consider the strong electrolytes
$Z_{m}X_{n},U_{m}Y_{p}$ and $V_{m}X_{n}$. Limiting molar conductivity
$(\Lambda{^\circ})$ of $U_{m}Y_{p}$ and $V_{m}X_{n}$ are 250 and
$440\ S\ cm^{2}\ mol^{- 1}$, respectively. The value of $(m + n + p)$ is
\_\_\_\_\_\_\_\_.
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 Ion $$Z^{n +}$$ $$U^{p +}$$ $$V^{n +}$$ $$X^{m -}$$ $$Y^{m -}$$
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 $$\lambda{^\circ}\left( S\ cm^{2}\ mol^{- 1} \right)$$ 50.0 25.0 100.0 80.0 100.0
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Answer: (7)
Solution: $Z_{m}X_{n},$ $U_{m}Y_{p},$ $V_{m}X_{n}$
$\lambda_{M}^{\infty} = \ ?$ 250 $440\ S\ cm^{2}mol^{- 1}$
$$\lambda_{M}^{\infty}\left( U_{m}Y_{p} \right) = m \times 25 + p(100) = 250$$
$25m + 100p = 250$ ... (i)
$\lambda_{M}^{\infty}\left( V_{m}X_{m} \right) = m \times 100 + n \times 80 = 440$
... (ii)
For
$Z_{m}X_{n} \Rightarrow \lambda_{M}^{c} = \lambda_{M}^{\infty} - b\sqrt{c}$
⇒ $\lambda_{n}^{c} = 340$ (From graph)
$\lambda_{M}^{\infty}\left( Z_{m}X_{n} \right) = m \times 50 + n \times 80 = 340$
... (iii)
From equation (ii) & (iii) $m = 2$ & $+ p = 2;n =$`<!-- -->`{=html}3
$$(m + n + p) = 7$$